# Homework Help: Gravitation force of a disk with a hole

1. May 4, 2017

### toony12362

1. The problem statement, all variables and given/known data
Hello, I have derived the equation for the gravitational force for a disk to be
2Ggm/a^2(1-x/sqrt(a^2 - x^2) when an object is added on top of the system. My question is would the force still be somewhat similar if the disk now had a small hollow of radius c and from the center to the end of the disk the distance is d?

3. The attempt at a solution
Im assuming it will be and that the only difference I would make is that my radius is now (d-c) or my new area would now be da = 2 * pi * (d^2 - c^2) but im not sure if that logic makes sense.

2. May 4, 2017

### Staff: Mentor

A sketch or at least a description of the variables would help.
The expression has mismatching brackets.

A disk with an empty interior can be described as sum of a larger disk with positive mass and a smaller disk with negative mass.

3. May 5, 2017

### toony12362

Sorry! Here is a sketch. Im assumning if we have a regular disk then the gravitional force can be modeled by the force equation in the diagram!

Now if we have a disk with an empty interior of radius b like this figure

Would the only thing that change in my equation would now be the radius? That the new radius would be (a -b) or would it be my area such that A = pi*b^2 - pi * a^2 or A=pi * (b-a)^2

4. May 5, 2017

### haruspex

From your second post, I gather that is:
$\frac{2Ggm}{a^2}\left(1-\frac x{\sqrt{a^2 - x^2}}\right)$
But that cannot be right. When x=0 the force should be 0.

5. May 5, 2017

### Staff: Mentor

@haruspex: The force is discontinuous at x=0 for an ideal disk, the formula is for x>0.
No, but you can model the force as sum of two disks, one with a positive and one with a negative mass.

The square root in your formula cannot be correct - it becomes undefined for x>a.

6. May 9, 2017